Why can't O form OF6?

Question

At first I thought that the reason is what I was taught, the absence of vacant d-orbitals but then I did some research. Most of the answers stated what I initially thought but one particular answer told that it's a common misconception (d-orbital theory). According to it, there can be two possible reasons-

Not much electronegativity difference between O and F.
The atomic size of Oxygen is not sufficient for it to hold 6 atoms.

So which one is likely to be more appropriate, the d-orbital theory or the electronegativity and atomic size theory?

Specifically this answer: https://chemistry.stackexchange.com/a/100938 — Karsten, Sep 11 '19 at 10:11
While "OF6" looks pretty absurd, there's possibility of existence of OF3+ cation which would have IV+ ox. state for oxygen! — Mithoron, Sep 11 '19 at 14:02
Whose to say it can't? Maybe we just need super low temperatures and gigapascals or terapascals of pressure.... — Curt F., Sep 11 '19 at 16:46
@mithoron wow! Inquiring minds want to know your references! — Oscar Lanzi, Sep 11 '19 at 22:34
@OscarLanzi https://www.sciencedirect.com/science/article/abs/pii/S0022113999001396 — Nilay Ghosh, Sep 12 '19 at 01:55
https://chemistry.stackexchange.com/questions/23496/how-does-absence-of-nd-orbital-in-oxygen-affect-its-valency — Mithoron, Sep 13 '19 at 16:51

Oscar Lanzi · Accepted Answer · 2021-03-11T02:04:17.757

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One possible explanation involves atomic size. You need a large enough atom in the middle to stabilize all the occupied orbitals in the octahedral structure; sulfur makes it but oxygen does not.

This reference gives the molecular orbitals for sulfur hexafluoride. A key feature of these molecular orbitals is a doubly degenerate pair, labeled $\mathrm{2e_g}$ in the above source, which effectively contains the "extra" valence electrons in fluorine-based orbitals.

The picture below, taken from a different source due to limited downloading options from the free one, includes one of these combinations (here labeled $\mathrm{3e_g}$, as different nomenclature is used) among others.

enter image description here

If we look carefully we see that the ostensibly nonbonding orbital has antibonding overlap between adjacent lobes and so is technically antibonding. The same is true of the other component of this degenerate pair. So to make the orbital effectively nonbonding and keep the octahedral structure from blowing apart, we need a large central atom to act as a "spacer" between orbital lobes. Apparently sulfur is large enough, oxygen is not.

edited Mar 11 '21 at 02:04

answered Sep 11 '19 at 10:04

Oscar Lanzi

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Nice answer, but what is "cathedral arructure"? I assume a typo, but not sure what you meant! – gilleain Sep 11 '19 at 11:29
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Corected, I think. Tiny keys ơn my device + fat fingers + overeager autocorrect = huge struggles for me to type. Is there a stylus for Galaxy A50 phones? – Oscar Lanzi Sep 11 '19 at 12:56
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@OscarLanzi Sorry for the off-topic comment, but I see you complaining about problems with typing and formatting quite often. The problem is, pretty much every smartphone is a device intended for consumption, not creation. All the external wireless or on-board keyboards I tried were horrible for anything remotely productive. If you would ask me, you were better off buying a refurbished Thinkpad (X series if you need something compact) for about $100, use smartphone as a hotspot and have a superior typing experience. – andselisk Mar 10 '21 at 19:23

Why can't O form OF6?

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